# 17.2 - A Triangular Support

17.2 - A Triangular Support

We now have many of the gory definitions behind us. One of the definitions we learned in particular is that two random variables $$X$$ and $$Y$$ are independent if and only if:

$$P(X=x, Y=x)=P(X=x)\times P(Y=y)$$

for all $$x\in S_1, y\in S_2$$. Otherwise, $$X$$ and $$Y$$ are said to be dependent. On the previous page, our example comprised two random variables $$X$$ and $$Y$$, which were deemed to be independent. On this page, we'll explore, by way of another example, two random variables $$X$$ and $$Y$$, which are deemed to be dependent.

## Example 17-2

Consider the following joint probability mass function:

$$f(x,y)=\dfrac{xy^2}{13}$$

in which the support is $$S=\{(x, y)\}=\{(1, 1), (1, 2), (2,2)\}$$. Are the random variables $$X$$ and $$Y$$ independent?

#### Solution

We are given the joint probability mass function as a formula. We can therefore easily calculate the joint probabilities for each $$(x, y)$$ in the support $$S$$:

when $$x=1$$ and $$y=1$$: $$f(1,1)=\dfrac{(1)(1)^2}{13}=\dfrac{1}{13}$$

when $$x=1$$ and $$y=2$$: $$f(1,2)=\dfrac{(1)(2)^2}{13}=\dfrac{4}{13}$$

when $$x=2$$ and $$y=2$$: $$f(2,2)=\dfrac{(2)(2)^2}{13}=\dfrac{8}{13}$$

Now that we have calculated each of the joint probabilities, we can alternatively present the p.m.f. in tabular form, complete with the marginal p.m.f.s of $$X$$ and $$Y$$, as:

As an aside, you should note that the joint support $$S$$ of $$X$$ and $$Y$$ is what we call a "triangular support," because, well, it's shaped like a triangle:

Anyway, perhaps it is easy now to see that $$X$$ and $$Y$$ are dependent, because, for example:

$$f(1,2)=\dfrac{4}{13} \neq f_X(1)\cdot f_Y(2)=\dfrac{5}{13} \times \dfrac{12}{13}$$

Note though that, in general, any two random variables $$X$$ and $$Y$$ having a joint triangular support must be dependent because you can always find:

$$f(x)\times f(y)=c\ne0$$

for some non-zero constant $$c$$. For example, for the joint p.m.f. above:

$$f_X(2)\times f_Y(1)=\left(\frac{8}{13}\right)\times\left(\frac{1}{13}\right)=\frac{8}{169}\ne 0=f_{X,Y}(2,1)$$

In general, random variables with rectangular support may or may not be independent.

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